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(D) According to the Stefan-Boltzmann law,the rate of heat loss by the object is given by $P = \sigma A T^4$.Since the container is at $0 \ K$,the net

Vedclass · Accepted Answer

$9$

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(D) According to the Stefan-Boltzmann law,the rate of heat loss by the object is given by $P = \sigma A T^4$.Since the container is at $0 \ K$,the net heat loss is $\sigma A T^4 = -ms \frac{dT}{dt}$,where $ms$ is the heat capacity $C$.Rearranging the terms: $\frac{dT}{T^4} = -\frac{\sigma A}{C} dt = -k dt$.Integrating from $T_i$ to $T_f$: $\int_{T_i}^{T_f} T^{-4} dT = -k \int_0^t dt$.This gives $\left[ \frac{T^{-3}}{-3} \right]_{T_i}^{T_f} = -kt$,or $\frac{1}{3} \left( \frac{1}{T_f^3} - \frac{1}{T_i^3} \right) = kt$.For $t_1$: $kt_1 = \frac{1}{3} \left( \frac{1}{100^3} - \frac{1}{200^3} \right) = \frac{1}{3 \cdot 100^3} \left( 1 - \frac{1}{8} \right) = \frac{7}{3 \cdot 100^3 \cdot 8} = \frac{7}{24 \cdot 10^6}$.For $t_2$: $kt_2 = \frac{1}{3} \left( \frac{1}{50^3} - \frac{1}{200^3} \right) = \frac{1}{3 \cdot 50^3} \left( 1 - \frac{1}{64} \right) = \frac{63}{3 \cdot 50^3 \cdot 64} = \frac{63}{3 \cdot 125000 \cdot 64} = \frac{63}{24 \cdot 10^6}$.Therefore,$\frac{t_2}{t_1} = \frac{63}{7} = 9$.

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